A moving block having mass m collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be (a) 0.5 (b) 0.25 (c) 0.4 (d) 0.8

Coefficient of Restitution (e) –

e = (v2 – v1) / (u2 – u1)

Ratio of relative velocity after collision to relative velocity before collision

Let velocity of 4m mass after collision be v

Then,

mv = 4mv

v = v/4v

e = (velocity of separation)/(velocity of approach)

e = v/4/v

e = ¼

We get,

e = 0.25

Hence, the correct answer is option (b)

Was this answer helpful?

  
   

0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question